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Palladium Algebra

Section 4.1 Introduction to Exponent Rules

In this section, we look at some properties of exponents that we can use when simplifying expressions that have multiplication and exponents.
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Figure 4.1.1. Alternative Video Lesson
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Subsection 4.1.1 Exponent Basics

Before we discuss any exponent rules, let’s remind ourselves about exponent fundamentals. When working with expressions with exponents, we have the following vocabulary:
\begin{equation*} \text{base}^{\text{exponent}} = \text{power} \end{equation*}
For example, when we calculate \(8^{2} = 64\text{,}\) the base is \(8\text{,}\) the exponent is \(2\text{,}\) and the expression \(8^{2}\) is called the 2nd power of \(8\text{.}\)
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Exponents indicate repeated multiplication. When the exponent is a positive integer, the power can be rewritten as repeated multiplication of the base. For example, the \(4\)th power of \(3\) can be written as \(4\) factors of \(3\) like so:
\begin{equation*} 3^{4} = 3 \cdot 3 \cdot 3 \cdot 3 \end{equation*}
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Subsection 4.1.2 Exponent Rules

Product Rule.

If we write out \(3^5\cdot 3^2\) without using exponents, we’d have:
\begin{equation*} 3^5 \cdot 3^2 = \left(3 \cdot 3\cdot 3\cdot 3\cdot 3\right) \cdot \left(3 \cdot 3\right) \end{equation*}
If we then count how many \(3\)s are being multiplied together, we find we have \(5+2=7\text{,}\) a total of seven \(3\)s. So \(3^5\cdot 3^2\) simplifies like this:
\begin{align*} 3^5\cdot 3^2 \amp= 3^{5+2}\\ \amp= 3^7 \end{align*}
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Example 4.1.2.

Simplify \(x^2\cdot x^3\text{.}\)
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To simplify \(x^2\cdot x^3\text{,}\) we write this out in its expanded form, as a product of \(x\)’s, we have
\begin{align*} x^2\cdot x^3 \amp=(x\cdot x)(x \cdot x \cdot x)\\ \amp=x\cdot x\cdot x \cdot x \cdot x\\ \amp=x^5 \end{align*}
Note that we obtained the exponent of \(5\) by adding \(2\) and \(3\text{.}\)
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This demonstrates our first exponent rule, the Product Rule: when multiplying two expressions that have the same base, we can simplify the product by adding the exponents.
\begin{equation} x^m \cdot x^n = x^{m+n}\tag{4.1.1} \end{equation}
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Checkpoint 4.1.3.

Use the properties of exponents to simplify the expression.
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\({t^{19}}\cdot{t^{2}}\)
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Explanation.
We add the exponents as follows:
\begin{equation*} \begin{aligned} t^{19}\cdot t^{2}\amp =t^{19+2}\\ \amp =t^{21} \end{aligned} \end{equation*}
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Recall that \(x=x^1\text{.}\) It helps to remember this when multiplying certain expressions together.
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Example 4.1.4.

Multiply \(x(x^3+2)\) by using the distributive property.
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According to the distributive property,
\begin{equation*} x(x^3+2)=x\cdot x^3 + x\cdot2 \end{equation*}
How can we simplify that term \(x\cdot x^3\text{?}\) It’s really the same as \(x^1\cdot x^3\text{,}\) so according to the Product Rule, it is \(x^4\text{.}\) So we have:
\begin{align*} x(x^3+2)\amp=x\cdot x^3 + x\cdot2\\ \amp=x^4+2x \end{align*}
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Power to a Power Rule.

If we write out \(\left(3^5\right)^2\) without using exponents, we’d have \(3^5\) multiplied by itself:
\begin{align*} \left(3^5\right)^2 \amp= \left(3^5\right)\cdot \left(3^5\right)\\ \amp= \left(3\cdot 3\cdot 3\cdot 3 \cdot 3 \right) \cdot \left(3 \cdot 3\cdot 3\cdot 3\cdot 3\right) \end{align*}
If we again count how many \(3\)s are being multiplied, we have a total of two groups each with five \(3\)s. So we’d have \(2\cdot 5=10\) instances of a \(3\text{.}\) So \(\left(3^5\right)^2\) simplifies like this:
\begin{align*} \left(3^5\right)^2 \amp= 3^{2\cdot 5}\\ \amp= 3^{10} \end{align*}
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Example 4.1.5.

Simplify \(\left(x^2\right)^3\text{.}\)
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To simplify \(\left(x^2\right)^3\text{,}\) we write this out in its expanded form, as a product of \(x\)’s, we have
\begin{align*} \left(x^2\right)^3 \amp=\left(x^2\right) \cdot \left(x^2\right)\cdot\left(x^2\right)\\ \amp=(x \cdot x)\cdot (x \cdot x)\cdot (x \cdot x)\\ \amp=x^6 \end{align*}
Note that we obtained the exponent of \(6\) by multiplying \(2\) and \(3\text{.}\)
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This demonstrates our second exponent rule, the Power to a Power Rule: when a base is raised to an exponent and that expression is raised to another exponent, we multiply the exponents.
\begin{equation*} \left(x^m\right)^n = x^{m \cdot n} \end{equation*}
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Checkpoint 4.1.6.

Use the properties of exponents to simplify the expression.
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\(\displaystyle{\left(y^{2}\right)^{9}}\)
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Explanation.
We multiply the exponents as follows:
\begin{equation*} \begin{aligned} \left(y^{2}\right)^{9}\amp =y^{2\cdot9}\\ \amp =y^{18} \end{aligned} \end{equation*}
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Product to a Power Rule.

The third exponent rule deals with having multiplication inside a set of parentheses and an exponent outside the parentheses. If we write out \(\left(3t\right)^5\) without using an exponent, we’d have \(3t\) multiplied by itself five times:
\begin{equation*} (3t)^5= (3t)(3t)(3t)(3t)(3t) \end{equation*}
Keeping in mind that there is multiplication between every \(3\) and \(t\text{,}\) and multiplication between all of the parentheses pairs, we can reorder and regroup the factors:
\begin{align*} \left(3t\right)^5 \amp= (3\cdot t)\cdot (3\cdot t)\cdot (3\cdot t)\cdot (3\cdot t)\cdot (3\cdot t)\\ \amp= \left(3\cdot 3\cdot 3\cdot 3\cdot 3 \right) \cdot \left(t \cdot t \cdot t \cdot t \cdot t\right)\\ \amp= 3^5 t^5 \end{align*}
We could leave it written this way if \(3^5\) feels especially large. But if you are able to evaluate \(3^5=243\text{,}\) then perhaps a better final version of this expression is \(243t^5\text{.}\)
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We essentially applied the outer exponent to each factor inside the parentheses. It is important to see how the exponent \(5\) applied to both the \(3\) and the \(t\text{,}\) not just to the \(t\text{.}\)
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Example 4.1.7.

Simplify \((xy)^5\text{.}\)
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To simplify \((xy)^5\text{,}\) we write this out in its expanded form, as a product of \(x\)’s and \(y\)’s, we have
\begin{align*} (xy)^5 \amp=(x \cdot y) \cdot (x \cdot y) \cdot (x \cdot y) \cdot (x \cdot y) \cdot (x \cdot y)\\ \amp=(x \cdot x \cdot x \cdot x \cdot x) \cdot (y \cdot y \cdot y \cdot y \cdot y)\\ \amp=x^5 y^5 \end{align*}
Note that the exponent on \(xy\) can simply be applied to both \(x\) and \(y\text{.}\)
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This demonstrates our third exponent rule, the Product to a Power Rule: when a product is raised to an exponent, we can apply the exponent to each factor in the product.
\begin{equation*} \left(x\cdot y\right)^n = x^{n}\cdot y^{n} \end{equation*}
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Checkpoint 4.1.8.

Use the properties of exponents to simplify the expression.
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\(\displaystyle{\left(3t\right)^2}\)
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Explanation.
We multiply the exponents and apply the rule \((ab)^m=a^m\cdot b^m\) as follows:
\begin{equation*} \begin{aligned} \left(3t\right)^{2}\amp =(3)^{2}t^{2}\\ \amp =9t^{2} \end{aligned} \end{equation*}
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List 4.1.9. Summary of the Rules of Exponents for Multiplication
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If \(b\) and \(c\) are real numbers, and \(m\) and \(n\) are positive integers, then we have the following rules:
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Product Rule
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\(\displaystyle b^{m} \cdot b^{n} = b^{m+n}\)
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Power to a Power Rule
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\(\displaystyle (b^{m})^{n} = b^{m\cdot n}\)
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Product to a Power Rule
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\(\displaystyle (bc)^{m} = b^{m} \cdot c^{m}\)
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Many examples will make use of more than one exponent rule. In deciding which exponent rule to work with first, it’s important to remember that the order of operations still applies.
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Example 4.1.10.

Simplify the following expressions.
  1. \(\displaystyle \left(3^7r^5\right)^4\)
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  2. \(\displaystyle \left(t^3\right)^2\cdot \left(t^4\right)^5\)
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Explanation.
  1. Since we cannot simplify anything inside the parentheses, we’ll begin simplifying this expression using the Product to a Power Rule. We’ll apply the outer exponent of 4 to each factor inside the parentheses. Then we’ll use the Power to a Power Rule to finish the simplification process.
    \begin{align*} \left(3^7r^5\right)^4 \amp= \left(3^7\right)^4 \cdot \left(r^5\right)^4\\ \amp= 3^{7\cdot4} \cdot r^{5\cdot 4}\\ \amp= 3^{28}r^{20} \end{align*}
    Note that \(3^{28}\) is too large to actually compute, even with a calculator, so we leave it written as \(3^{28}\text{.}\)
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  2. According to the order of operations, we should first simplify any exponents before carrying out any multiplication. Therefore, we’ll begin simplifying this by applying the Power to a Power Rule and then finish using the Product Rule.
    \begin{align*} \left(t^3\right)^2\cdot \left(t^4\right)^5 \amp= t^{3\cdot2}\cdot t^{4\cdot5}\\ \amp= t^6 \cdot t^{20}\\ \amp= t^{6+20}\\ \amp= t^{26} \end{align*}
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Remark 4.1.11.

We cannot simplify an expression like \(x^2y^3\) using the Product Rule, as the factors \(x^2\) and \(y^3\) do not have the same base.
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Reading Questions 4.1.3 Reading Questions

1.

How many exponent rules are discussed in this section? Write an example of each rule in action.
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2.

The order of operations say that operations inside parentheses should get the highest priority. But with \((5x)^3\text{,}\) you cannot actually do anything with the \(5\) and the \(x\text{.}\) Which exponent rule allows you to sidestep the order of operations and still simplify this expression a little?
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Exercises 4.1.4 Exercises

Review and Warmup

Exercise Group.
Evaluate each power expression.
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3.
(a)
\(\left({{\frac{3}{2}}}\right)^2\)
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(b)
\(\left({-{\frac{1}{5}}}\right)^4\)
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(c)
\(\left({-{\frac{1}{3}}}\right)^5\)
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(d)
\(-\left({{\frac{3}{2}}}\right)^4\)
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4.
(a)
\(\left({{\frac{3}{4}}}\right)^5\)
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(b)
\(\left({-{\frac{2}{3}}}\right)^4\)
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(c)
\(\left({-{\frac{5}{4}}}\right)^3\)
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(d)
\(-\left({{\frac{3}{4}}}\right)^4\)
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Skills Practice

Exercise Group.
Use the properties of exponents to simplify the expression.
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15.
\({\left(8^{3}\right)^{2}}\)
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16.
\({\left(9^{5}\right)^{8}}\)
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17.
\({\left(a^{10}\right)^{5}}\)
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18.
\({\left(c^{5}\right)^{2}}\)
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23.
\({\left(3r^{6}\right)^{3}}\)
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24.
\({\left(5u^{3}\right)^{4}}\)
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27.
\({\left(-2c^{5}\right)^{2}}\)
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28.
\({\left(-3f^{3}\right)^{3}}\)
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29.
\({\left(3h\right)^{4}\mathopen{}\left(9h\right)^{6}\mathopen{}\left(8h\right)^{10}}\)
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30.
\({\left(9k\right)^{4}\mathopen{}\left(4k\right)^{10}\mathopen{}\left(6k\right)^{9}}\)
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31.
\({-\left(4n^{4}\right)^{4}}\)
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32.
\({-\left(2r^{2}\right)^{2}}\)
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33.
\({\frac{u^{7}}{8}\frac{u^{3}}{6}}\)
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34.
\({\frac{x^{5}}{5}\frac{x^{3}}{9}}\)
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35.
\({\left(\frac{z^{5}}{2}\right)^{3}}\)
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36.
\({\left(\frac{c^{8}}{2}\right)^{9}}\)
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Exercise Group.
Use the properties of exponents together with the distributive property to write the expression in a new simplified way where there are no grouping symbols.
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37.
\({-3e\mathopen{}\left(6e+5\right)}\)
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38.
\({-9h\mathopen{}\left(5h-4\right)}\)
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39.
\({-9k^{7}\mathopen{}\left(8k-6\right)}\)
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40.
\({2n^{4}\mathopen{}\left(7n-9\right)}\)
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41.
\({-9r^{2}\mathopen{}\left(2r^{9}+9r^{6}\right)}\)
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42.
\({-9u^{7}\mathopen{}\left(9u^{4}-10u^{3}\right)}\)
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Comparisons.
Simplify each expression, taking note of the different behavior when adding versus multiplying. If it’s not possible to simplify an expression more than it already is, then the answer should just be the same given expression.
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Simplifying Expressions.
Simplify the given expression.
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53.
\({5z^{4}\mathopen{}\left(3z^{3}\right)^{2}}\)
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54.
\({5c^{3}\mathopen{}\left(4c^{2}\right)^{4}}\)
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55.
\({4e^{7}f^{6}\mathopen{}\left(2e^{9}f^{3}\right)^{2}}\)
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56.
\({2h^{2}w^{5}\mathopen{}\left(4h^{9}w^{8}\right)^{2}}\)
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57.
\({-8j\cdot 4j^{6}-5j^{5}\cdot 2j^{2}}\)
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58.
\({8n\cdot 7n^{8}-3n^{2}\cdot 6n^{7}}\)
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59.
\({-7q^{4}\cdot 4q^{8}+2q^{5}\cdot 5q^{7}}\)
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60.
\({7u^{2}\cdot 8u^{8}-2u^{5}\cdot 4u^{5}}\)
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61.
\({5x\mathopen{}\left(-8x^{9}\right)^{3}+\left(8x^{4}\right)^{2}\cdot 2x^{20}}\)
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62.
\({6z^{5}\mathopen{}\left(2z^{7}\right)^{3}-\left(-6z^{6}\right)^{2}\cdot 9z^{14}}\)
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Applications

63.
PTX:ERROR: WeBWorK problem Palladium_Algebra/4_1_4_63.pg with seed 1771 is either empty or failed to compile Use -a to halt with full PG and returned content
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64.
PTX:ERROR: WeBWorK problem Palladium_Algebra/4_1_4_64.pg with seed 1772 is either empty or failed to compile Use -a to halt with full PG and returned content
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65.
There are about \(10^{19}\) grains of sand on Earth. On average there are also about \(10^{19}\) atoms in a grain of sand. Using these numbers, how many total atoms are there on Earth in grains of sand?
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66.
There are roughly \(10^8\) stars in an average galaxy, and there are roughly \(10^{12}\) galaxies in the observable universe. Using these numbers, how many stars are there in the observable universe? Answer using a simplified exponent expression, not the actual result.
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Challenge

67.
(a)
Let \(x^{10}\cdot x^{a}=x^{20}\text{,}\) where \(a\) is a natural number. How many possibilities are there for \(a\text{?}\)
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(b)
Let \(x^{b}\cdot x^{c}=x^{90}\text{,}\) where \(b\) and \(c\) are natural numbers. How many possibilities are there for \(b\text{?}\)
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(c)
Let \(x^{d}\cdot x^{e}=x^{1000}\) where \(d\) and \(e\) are natural numbers. How many possibilities are there for \(d\text{?}\)
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69.
Fill in the blanks with algebraic expressions that make the equation true. You may not use \(0\) or \(1\) in any of the blank spaces. For example, if \(\mathord{?}+\mathord{?}=8x\text{,}\) then one possible answer is \(3x+5x=8x\text{.}\)
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There are infinitely many correct answers. Be creative. After finding a correct answer, see if you can come up with a different answer that is also correct.
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  1. \({}+{}\)\({}={-11x}\)
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  2. \({}+{}\)\({}={-11x^{25}}\)
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  3. \({}\cdot{}\)\({}\cdot{}\)\({}={5x^{50}}\)
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